Let $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$. We observe the symmetry of the function over the interval $[-1, 1]$.
Recall that $f(x)$ is an odd function if $f(-x) = -f(x)$. Let us evaluate $f(-x)$:
$$f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} = \frac{-x^3}{x^2 + 2|x| + 1} = -f(x)$$Since $f(-x) = -f(x)$, the function is an odd function.
For an odd function $f(x)$, the property of definite integrals states that:
$$\int_{-a}^{a} f(x) dx = 0$$Applying this to our integral where $a = 1$:
$$\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx = 0$$Final Answer: 0
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